Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 317 / 3 / ii

Solution

\frac{d P _{rad}}{d r} = - \frac{κ ρ L _{r}}{4 π c r ^{2}} .

\frac{d P _{rad}}{d P} = \frac{κ L _{r}}{4 π c G m _{r}} = \frac{νL}{4 π c GM},

which is constant by assumption. Since both

P_{rad}

and

P

vanish at the surface, integration gives

P_{rad} = (1 - β) P

with constant

β

Eliminating

T

P = K ρ^{4/3}, K = [\frac{3}{a} (\frac{R}{μ})^{4} \frac{1 - β}{β ^{4}}]^{1/3} .

Thus the star is an

n = 3

ρ = ρ_{c} θ^{3}, r = α ξ, α^{2} = \frac{K}{π G} ρ_{c}^{- 2/3} .

\frac{1}{ξ ^{2}} \frac{d}{d ξ} (ξ^{2} \frac{d θ}{d ξ}) = - θ^{3}, θ (0) = 1, θ^{'} (0) = 0 .

The surface is the first zero

ξ_{1} ≃ 6.89685

ω_{3} = - ξ_{1}^{2} θ^{'} (ξ_{1}) ≃ 2.01824,

M = 4 π α^{3} ρ_{c} ω_{3} = \frac{4 ω _{3}}{π} (\frac{K}{G})^{3/2} .

Therefore

M = λ \frac{( 1 - β ) ^{1/2}}{β ^{2}},

where

λ = \frac{4 ω _{3}}{π} (\frac{3}{a})^{1/2} (\frac{R}{μ})^{2} G^{- 3/2} .

The additional radiative-equilibrium relation is

1 - β = νL / (4 π c GM)

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